Physics, 14.09.2019 06:30 queentaryn13

Apoint charge of 3 µc is located at x = -3.0 cm, and a second point charge of -10 µc is located at x = +4.0 cm. where should a third charge of +6.0 µc be placed so that the electric field at x = 0 is zero?

Answers: 1

Show answers

Answers

Answer from: Djbrace3278

The charge q₃ must be placed at X = +2.5 cm

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1µC= 10⁻6 C

1cm= 10⁻² m

Data

k = 8.99*10⁹ N×m²/C²

q₁ =+3 µC =3*10⁻⁶ C

q₂ = -10 µC =-10*10⁻⁶ C

q₃= +6µC =+6*10⁻⁶ C

d₁ = 3cm =3×10⁻² m

d₂ = 4cm = 4×10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

E₁:Field at point P due to charge q₁. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

E₂: Field at point P due to charge q₂. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

Problem development

E₃: Field at point P due to charge q₃. As the charge q₃ is positive, the field leaves the charge.

The direction of E₃ must be (- x) so that the electric field can be equal to zero at point P since E₁ and E₂ are positive, then, q₃must be located to the right of point P.

We make the algebraic sum of fields at point P due to the charges q1, q2, and q3:

E₁+E₂-E₃=0

$\frac{k*q_{1} }{d_{1}^{2} } +\frac{k*q_{2} }{d_{2}^{2} } -\frac{k*q_{3} }{d_{3}^{2} } =0$

We eliminate k

$\frac{q_{1} }{d_{1} ^{2} } +\frac{q_{2} }{d_{2} ^{2} }+\frac{q_{3} }{d_{3} ^{2} }=0$

We replace data

$\frac{3*10^{-6} }{(3*10^{-2})^{2} } +\frac{10*10^{-6} }{(4*10^{-2})^{2} } +\frac{6*10^{-6} }{d_{3} ^{2} } =0$

we eliminate 10⁻⁶

$\frac{3}{9*10^{-4} } +\frac{10}{16*10^{-4} } =\frac{6}{d_{3}^{2} }$

$(\frac{1}{10^{-4} }) *(\frac{1}{3} +\frac{5}{8}) =\frac{6}{d_{3}^{2} }$

$\frac{23*10^{4} }{24} =\frac{6}{d_{3} ^{2} }$

$d_{3} =\sqrt{\frac{6*24}{23*10^{4} } }$

$d_{3} =2.5*10^{-2} m\\d_{3} =2.5 cm$

The charge q₃ must be placed at X = +2.5 cm

Apoint charge of 3 µc is located at x = -3.0 cm, and a second point charge of -10 µc is located at x

Answer from: Quest

okay i have so many questions. but the big one here is: um wt..f is this mess? ?

Answer from: Quest

answer: you might love summer weather, but you don’t want it to feel like the sun is inside your car. but a hot car that just won’t seem to cool down might be a bigger of a problem than you think. your car relies on being at the right temperature.

explanation:

Another question on Physics

Physics, 21.06.2019 20:30

If forces acting on an object are unbalanced, which factors may result fron an unbalanced force? check all that apply

Answers: 1

Answer

Physics, 21.06.2019 21:30

According to social exchange theory, altruistic behaviors benefit the individual who performs them. t/f

Answers: 1

Answer

Physics, 21.06.2019 22:00

What is the role of the ocular lens?

Answers: 1

Answer

Physics, 22.06.2019 02:30

Agas contained within a piston-cylinder assembly undergoes three processes in series: process 12: compression with pv= constant from 1 bar and 1 liter to 4 bar. process 23: constant pressure expansion to 1 liter. process 31: constant volume calculate the pressure and volume at each state, and sketch the processes on a p-vdiagram labeled with pressure and volume values at each numbered stat

Answers: 2

Answer

You know the right answer?

Apoint charge of 3 µc is located at x = -3.0 cm, and a second point charge of -10 µc is located at x...

Questions